A grating used at normal incidence gives a spectral line 6000 $\mathring{A}$ of a certain order that coincides with another color 4800 $\mathring{A}$ of the next higher order. If angle of diffraction is $30^0$ calculate the grating element.

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

λ$_1$= 6000 A° = 6 x 10$^{-5}$ cm, λ$_2$ = 4800 A° = 4.8 x 10$^{-5}$ cm, θ$_1$=θ$_2$ =30°, sinθ$_1$=sinθ$_2$=0.5

$(a+b) sinθ_1 = n_1 λ_1 $ ……………………………………(1)

$(a+b) sinθ_2 = (n_1 +1) λ_2 $ …………………………………(2)

Dividing equation 1 and 2,

$ n_1 λ_1 = (n_1 +1) λ_2 \\[2ex] n_1 λ_1 = n_1 λ_2 + λ_2 \\[2ex] n_1 ( λ_1 - λ_2 ) = λ_2 \\[2ex] n_1 = \frac{λ_2}{( λ_1 - λ_2 )} \\[2ex] n_1 = \frac{4.8 \times 10^{ -5}}{(6 \times 10^{-5} - 4.8 \times 10^{ -5} )} \\[2ex] n_1 = 4 $

Substituting in equation 1,

(a+b) sin30 = 4 x 6 x 10$^{-5}$
(a+b) = 48 x 10$^{-5}$ cm